Find an equation for the plane that passes through (2,4,-1) and contains the line of intersection of the planes x-y-4z=2 and -2x+y+2z=3

Question

I have no idea what Im doing wrong but I keep getting the wrong answer. What I did first was find two points. So I set z=0 and x=0 . Then I created two vectors and did the cross product for ABxAC to get the normal. But the final equation I got was wrong 

Answer 1

Note that the first given plane x - y - 4z = 2 already contains (2,4,-1) and so it is the answer to this problem.

 

 

I will outline the general procedure here for how to find such a plane in case you are given a version of the problem where neither plane given has the point.

 

You can just take any linear combination to get a plane containing the line. So all we have to do is find which linear combination will result in the plane that also contains (2,4,-1).

 

Say we add A times the first equation and B times the second.

 

A(x - y - 4z) + B(-2x + y + 2z) = 2A + 3B

 

A(2 - 4 - 4(-1)) + B(-2(2) + 4 + 2(-1)) = 2A + 3B

 

2A - 2B = 2A + 3B

 

-2B = 3B

 

B = 0

 

Thus any multiple of the equation x - y - 4z = 2 will be a valid answer.

 

If the fully simplified equation in A and B contains both of them then just pick any value for A and solve for the corresponding B.

 

If the equation only contains one of A and B, it indicates that one of the planes in the problem is the correct answer.

 

If both A and B get eliminated then either the point is on the line of intersection (all linear combinations of the plane equations are solutions), or the two planes are parallel (no solution).